Geometric proof that a sum $\sum{(1 + 2+ 3 + .... + n)}$ takes the form $\frac{(n+1)n}{2}$ I understand that the sum  $\sum{(1 + 2+ 3 + .... + n)}$ takes the form $\frac{(n+1)n}{2}$.
This can be shown symbolically: $1+2+3+...+n$, written backwards is $n+(n-1)+(n-2)+(n-3)+...+3+2+1$.  If you add these two identical values together, you get:
$$1+2+3\\ n+(n-1)+(n-2)$$
$n+1$; if you add the second term of each, you get $(n-1)+2=(n+1)$; the third sum is $(n-2)+3=(n+1)$.  Hence, $(n+1)$ $n$ times is double the value.
Before I understood this, however, I figured that there would be a $\frac{1}{2}$ in the term.  Take $3^2$ for example, expressed in an actual square:
$$1+1+1 \\ 1+1+1\\1+1+1$$
Now, we instead want to sum all integers from $1$ to $3$, we would have:
$$1\\\space\space\space\space\space\space1+1\\\space\space\space\space\space\space\space\space\space\space\space\space\space1+1+1$$
Of course a right triangle with sides of $n$ is $\frac{1}{2}n^2$.  However, the squares which we can use to makeup a larger square don't evenly fit in the triangle.
Does this summation lend itself to any geometric proof?
 A: Here's a geometric proof, shamelessly stolen from here:
This shows that two copies of $10 = \frac{4(4+1)}{2}$ combine to make a $4\times 5$ rectangle.
A: It does actually. Take the example you posted. You have a right triangle with legs of length $n$. As you said, this triangle has $\sum_{i=1}^{n} i$ squares in it. When you duplicate this triangle and flip it and attach it back diagonally to the original triangle, you get a rectangle of sides $n$ and $n+1$. The area of this is equal to $n(n+1)$, but it is also equal to $2\sum_{i=1}^{n} i$. Therefore $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}$.
A: Take a look at this image:

The height and width of this object is $n$. If we draw a line from the center top all the way to the bottom center of the left-most square, then do the same for the right and connect the two bottom centers, we get a triangle of area $\frac{n^2}{2}$. The small right triangles left each have an area of $\frac{1}{4}$, so the total area of all these tiny triangles is $\frac{n}{2}$ (the sum of the areas of the triangles on the left and then on the right). Then add the area of the big triangle and the area of the small triangles to get the answer.
